Automorph-conjugacy is transitive: Difference between revisions

Revision as of 16:17, 19 December 2014

This article gives the statement, and possibly proof, of a subgroup property (i.e., automorph-conjugate subgroup) satisfying a subgroup metaproperty (i.e., transitive subgroup property)
View all subgroup metaproperty satisfactions | View all subgroup metaproperty dissatisfactions |Get help on looking up metaproperty (dis)satisfactions for subgroup properties
Get more facts about automorph-conjugate subgroup |Get facts that use property satisfaction of automorph-conjugate subgroup | Get facts that use property satisfaction of automorph-conjugate subgroup|Get more facts about transitive subgroup property

Statement

Suppose $H \leq K \leq G$ are groups such that $H$ is an automorph-conjugate subgroup of $K$ , and $K$ is an automorph-conjugate subgroup of $G$ . Then, $H$ is an automorph-conjugate subgroup of $G$ .

Proof

Given: $H$ is an automorph-conjugate subgroup of $K$ , and $K$ is an automorph-conjugate subgroup of $G$ . An automorphism $σ$ of $G$ .

To prove: There exists $x \in G$ such that $σ (H) = x H x^{- 1}$

Proof: Clearly, $σ (H) \leq σ (K)$ , and since $K$ is automorph-conjugate subgroup of $G$ , there exists $g \in G$ such that $g σ (K) g^{- 1} = K$ . Thus, $c_{g} \circ σ$ (conjugation by $g$ , composed with $σ$ ), gives an automorphism of $K$ . Since $H$ is automorph-conjugate inside $K$ , there exists $h \in K$ such that $g σ (H) g^{- 1} = h H h^{- 1}$ . Rearranging, we see that $σ (H) = g^{- 1} h H h^{- 1} g$ , so setting $x = g^{- 1} h$ works.

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 ==Proof==
-Suppose <math>H</math> is an automorph-conjugate subgroup of <math>K</math>, and <math>K</math> is an automorph-conjugate subgroup of <math>G</math>. We want to show that <math>H</math> is an automorph-conjugate subgroup of <math>G</math>.
+'''Given''': <math>H</math> is an automorph-conjugate subgroup of <math>K</math>, and <math>K</math> is an automorph-conjugate subgroup of <math>G</math>. An automorphism <math>\sigma</math> of <math>G</math>.
-For this, pick any automorphism <math>\sigma</math> of <math>H</math>. Clearly, <math>\sigma(H) \le \sigma(K)</math>, and since <math>K</math> is automorph-conjugate subgroup of <math>G</math>, there exists <math>g \in G</math> such that <math>g \sigma(K) g^{-1} = K</math>. Thus, <math>c_g \circ \sigma</math> (conjugation by <math>g</math>, composed with <math>\sigma</math>), gives an automorphism of <math>K</math>. Since <math>H</math> is automorph-conjugate inside <math>K</math>, there exists <math>h \in K</math> such that <math>g \sigma(H) g^{-1} = hHh^{-1}</math>. Rearranging, we see that <math>\sigma(H) = g^{-1}h H h^{-1}g</math>, a conjugate fo <math>H</math>.
+'''To prove''': There exists <math>x \in G</math> such that <math>\sigma(H) = xHx^{-1}</math>
+'''Proof''': Clearly, <math>\sigma(H) \le \sigma(K)</math>, and since <math>K</math> is automorph-conjugate subgroup of <math>G</math>, there exists <math>g \in G</math> such that <math>g \sigma(K) g^{-1} = K</math>. Thus, <math>c_g \circ \sigma</math> (conjugation by <math>g</math>, composed with <math>\sigma</math>), gives an automorphism of <math>K</math>. Since <math>H</math> is automorph-conjugate inside <math>K</math>, there exists <math>h \in K</math> such that <math>g \sigma(H) g^{-1} = hHh^{-1}</math>. Rearranging, we see that <math>\sigma(H) = g^{-1}h H h^{-1}g</math>, so setting <math>x = g^{-1}h</math> works.